IntroductionTheoryExperimentalData analysisWritten laboratory reportReferencesReferencesExperiment 4: Conductivity of electrolyte solutions(Dated: October 29, 2009)I. INTRODUCTIONPure water does not conduct electricity, but any solvated ionic species would contribute to conduction of electricity.An ionically conducting solution is called an electrolyte solution and the compound, which produces the ions as itdissolves, is called an electrolyte. A strong electrolyte is a compound that will completely dissociate into ions in water.Correspondingly, a weak electrolyte dissolves only partially. The conductivity of an electrolyte solution depends onconcentration of the ionic species and behaves differently for strong and weak electrolytes. In this work the electricconductivity of water containing various electrolytes will be studied. The data will be extrapolated to infinitelydilute solutions and the acidity constant for a given weak electrolyte will also be determined. Additional theoreticalbackground for electrolyte solutions can be found from Refs. [1–3].II. THEORYMovement of ions in water can be studied by installing a pair of electrodes into the liquid and by introducing apotential difference between the electrodes. Like metallic conducting materials, electrolyte solutions follow Ohm’slaw:R =UI(1)where R is the resistance (Ω, “ohms”), U is the potential difference (V, “Volts”), and I is the current (A, “Amperes”).Conductance G (S, Siemens or Ω−1) is then defined as reciprocal of the resistance:G =1R(2)Conductance of a given liquid sample decreases when the distance between the electrodes increases and increaseswhen the effective area of the electrodes increases. This is shown in the following relation:G = κAl(3)where κ is the conductivity (S m−1), A is the cross-sectional area of the electrodes (m2; e.g. the effective area availablefor conducting electrons through the liquid), and l is the distance between the electrodes (m). Molar conductivityΛm(S m2mol−1) is defined as:Λm=κc(4)where c is the molar concentration of the added electrolyte. A “typical” value for molar conductivity is 10 mS m2mol−1.The molar conductivity of an electrolyte would be independent of concentration if κ were proportional to theconcentration of the electrolyte. In practice, however, the molar conductivity is found to vary with the concentration(see Fig. 1). One reason for this variation is that the number of ions in the solution might not be proportional tothe concentration of the electrolyte. For example, the concentration of ions in a solution of a weak acid depends onthe concentration of the acid in a complicated way, and doubling the concentration of the acid does not double thenumber of ions. Another issue is that ions interact with each other and tend to slow down each other leading reducedconductivity. In this limit, the molar conductivity depends on square root of electrolyte molar concentration.In the 19th century Friedrich Kohlrausch discovered the following empirical relation between the molar concentrationof a strong electrolyte and the molar conductivity (Kohlrausch’s law) at low concentrations:Λm= Λ0m− K√c (5)Typeset by REVTEXFIG. 1: Variation of molar conductivity as a function of molar concentration. a) Strong electrolute and b) weak electrolyte.where K is a non-negative constant depending on the electrolyte and Λ0mis the limiting molar conductivity (e.g. themolar conductivity in the limit of zero concentration of the electrolyte). Furthermore, Kohlrausch was able to showthat Λ0mcan be expressed as a sum of contributions from its individual ions. If the limiting molar conductivity forthe cations is λ+and for the anions λ−, the “law of the independent migration of ions” states:Λ0m= v+λ++ v−λ−(6)where v+is the number of cations per formula unit, v is the corresponding number of anions, and λ+and λ−arethe limiting molar conductivities for cations and anions, respectively. For example, for HCl v+= 1 and v−= 1 butfor MgCl2we have v+= 1 and v = 2. Because weak electrolytes are not fully ionized in solution, the number ofions is not proportional to the concentration of the electrolyte but depends on the degree of dissociation (α). Theeffective molar conductivity can then be approximated in terms of α and the hypothetical molar conductivity of thefully ionized case (Λ0m):Λm= αΛ0m(7)When a weak acid dissociates in water solution, we have:HA(aq) + H2O(l) ⇋ H3O+(aq) + A−(aq) (8)The effective concentrations in solution are then given by (subscript 0 refers to the initial concentration of the acid):£H3O+¤= α [HA]0,£A−¤= α [HA]0, [HA] = (1 − α) [HA]0(9)The acidity constant (Ka) can now be written in terms of the ion activities (a):Ka=a¡H3O+¢a¡A−¢a (HA) a (H2O)|{z }= 1 (solvent)=a¡H3O+¢a¡A−¢a (HA)(10)In order to proceed, we write activity in an alternative form for each species (here i = H3O+, A−, HA, H2O):a(i) = γibibθ(11)2where γiis the activity coefficient (dimensionless) for species i, biis the molality for i (mol kg−1) and bθis the idealsolution molality (constant, 1 mol kg−1). Inserting Eq. (11) into Eq. (10) we get:Ka=γH3O+γA−γHA×γH3ObA−bHAbθ= Kγ× Kb(12)where notation of Kγand Kbare used for convenience. Do not confuse Kbwith the acidity constant! In dilute solutionsthe mean activity coefficient (γave=√γH3O+γA−; geometric mean value) can be calculated using the Debye-Hckellimiting law:log (γave) = −|zH3O+zA−|A√I (13)where z’s are the ionic charges, A is a constant (typically 0.509 for an aqueous solution at 25oC) and I is thedimensionless ionic strength of the solution given by:I =12NionsXi=1z2ibibθ(14)where Nionsis the number of difference ionic species in the solution. Note that log here denotes a logarithm withbase 10 (ln would denote the natural base logarithm). Since the activity coefficient for the neutral species (γHA) isequal to one and γ2ave= γH3O+γA−, the Eq. (12) gives:Ka= γ2ave× Kb(15)or by using logarithms:log (Ka) = 2 log (γave) + log (Kb) (16)In dilute solutions molalities are directly related to concentrations by:[i] = ci= biρ (17)where ciis the molar concentration of species i (mol L−1; usually denoted by species in brackets) and ρ is the densityof the solution (kg L−1). Inserting Eq. (17) into Eq. (16) we have:log (Ka) = 2 log (γave) + logãH3O+¤£A−¤ρbθ[HA]!(18)Note that the numerical value of ρbθis approximately